For the decolorization of `1` mol of `KMnO_4`, the moles of `H_2O_2` requiered are .

Options

A. ` 1//2`
B. ` 3//2`
C. ` 5//2`
D. `7//2`

Answer

C

Explanation

`overset(+7)(M)nO_(4)^(-) rarr overset(+2)(M)n^(2+)`
The oxidation number of Mn is decreasing from ` +7` to ` +2` Thus, 1 mol of `KMnO_4` contains `5` equivalents of ` KMnO_4`:
` H_2 overset (-1) (O_2)rarr overset(0)(O_(2))`
The oxidation number fo oxygen is increasing from ` -2 `to `0`. Thus, 1 mol of `H_2O_2` contains `2` equivalents of ` H_2O_2`. According to the law of equivalents, we need `5` equivalents of ` H_2O_2` to decolorize 1 mol of ` KMnO_4` Sicec `2` equivalents by ` H_2O_2` are provided by 1 mol , 5 equivalents will be provided by ` 5//2` mol. Alternatively, we can extract the answer by means of balanced equation :
`{:(2KMnO_(4)+3H_(2)SO_(4) rarr K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5[0]),(H_(2)O_(2)+[0] rarr H_(2)O+O_(2)"]"xx5),(bar(2KMnO_(4)+3H_(2)SO_(4)+5H_(2)O_(2) rarr K_(2)SO_(4)+2MnSO_(4)+8H_(2)O+5O_(2))):}`
According to equation, 2 mol of `KMnO_4` react with 5 mol of `H_2O_2`. Thus 1 mol of `KMnO_(4)` will react with `5//2` mol of `H_(2)O_(2)`.